Why is this sum equal to $0$? While solving a differential equation problem involving power series, I stumbled upon a sum (below) that seemed to be always equal to $0$, for any positive integer $s$.
$$
\sum_{k=0}^s \left( \frac{ \prod_{r=1}^k (-4r^2+10r-3) \prod_{r=1}^{s-k} (-4r^2+6r+1)}{2^s (2k)! (2s-2k+1)!} \times (2s-4k+1) \right)
$$
Why is this sum always equal to $0$?
The simplified version of this equation would be:
$$
\frac{1}{(-2)^s (2s+1)!} \sum_{k=0}^s \left( \binom{2s+1}{2k}  (2s-4k+1) \prod_{r=1}^k (4r^2-10r+3) \prod_{r=1}^{s-k} (4r^2-6r-1)  \right)
$$
or
$$
\frac{1}{(-2)^s (2s)!} \sum_{k=0}^s \left( \left( \binom{2s}{2k} - \binom{2s}{2k-1} \right) \prod_{r=1}^k (4r^2-10r+3) \prod_{r=1}^{s-k} (4r^2-6r-1)  \right)
$$
Update: 
$4r^2-10r+3$ and $4r^2-6r-1$ were derived from $n^2-5n+3$ by replacing $n$ with $2r$ and $2r+1$, respectively, which means this could be rewritten as:
$$
\frac{1}{(-2)^s (2s)!} \sum_{k=0}^s \left( \left( \binom{2s}{2k} - \binom{2s}{2k-1} \right) \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1}  \right)
$$
where $a_n=n^2-5n+3$
If the sum on the first line is indeed equal to $0$, then this would also be true:
$$
\sum_{k=0}^s \left( \binom{2s}{2k} \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right) = \sum_{k=0}^s \left( \binom{2s}{2k-1} \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right)
$$
Then, I changed the coefficients of $a_n$ into unknown constants to check whether this equality is true in other cases: $a_n=n^2-bn+c$
It seemed that this equality stands when $b$ is an odd integer greater than or equal to $5$ and $s$ is an integer greater than or equal to $(b-3)/2$; when these conditions are met, the value of $c$ does not seem to affect anything.
How could one mathematically derive or prove the above conclusion?
 A: Let 
$$
\alpha=\frac{5+\sqrt{13}}{4},\beta=\frac{5-\sqrt{13}}{4}
$$ be the roots of equation $4r^2-10r+3=0$, and 
$$
\gamma=\frac{3+\sqrt{13}}{4},\delta=\frac{3-\sqrt{13}}{4}
$$
be the roots of equation $4r^2-6r-1=0$. Note that $\alpha-\gamma=\beta-\delta=\frac{1}{2}$,$\ \alpha+\beta=\frac{5}{2}$ (this will be used below). Using the Pochhammer symbol 
$$
(x)_n=x(x+1) ... (x+n-1)
$$ 
and the identities
$$
(x)_{n-k}=(-1)^k\frac{(x)_n}{(-x-n+1)_k},
$$
$$
(2k)!=4^kk!\left(\frac{1}{2}\right)_k,\quad (2s-2k+1)!=2^{2s-2k}\left(1\right)_{s-k}\left(\frac{3}{2}\right)_{s-k}
$$
one can rewrite the sum as
$$
A(s)\cdot\sum_{k=0}^s\frac{(1-\alpha)_k(1-\beta)_k(-s)_k\left(-s-\frac{1}{2}\right)_k}{k!\left(\frac{1}{2}\right)_k(\gamma-s)_k(\delta-s)_k}\left(s-2k+\frac{1}{2}\right)
$$
where $A(s)$ is a non-essential constant that had absorbed all prefactors that depend only on $s$ (but not on $k$). Up to a prefactor this is equal to
$$
_7F_6\left({\textstyle a,1+\frac{a}{2},b,c,d,e,-s \atop \textstyle \frac{a}{2},1+a-b,1+a-c,1+a-d,1+a-e,1+a+s}\,;1\right)
$$
where $a=-s-\frac{1}{2}$,$\ b=1-\alpha$,$\ c=1-\beta$,$\ d=e=\frac{1+a}{2}$. One can check that $1+2a=b+c+d+e-s$, so the series can be summed using Dougall's formula
To show that this series under consideration is $0$ it is enough to note that (one can see this from the link above) this sum is proportional to the Pochhammer symbol $(1+a-b-c)_s=(1-s)_s$ which is $0$ for positive $s$.
